Solution 3.3:6c

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Before we even start thinking about transforming \displaystyle \log_2 and \displaystyle \log_3 to ln, we use the log laws

\displaystyle \begin{align}

\log a^b &= b\cdot\log a\,,\\[5pt] \log (a\cdot b) &= \log a+\log b\,, \end{align}

to simplify the expression

\displaystyle \begin{align}

\log_{3}\log _{2}3^{118} &= \log_{3}(118\cdot\log_{2}3)\\[5pt] &= \log_{3}118 + \log_{3}\log_{2}3\,\textrm{.} \end{align}

With help of the relation \displaystyle 2^{\log_{2}x} = x and \displaystyle 3^{\log_{3}x} = x and taking the natural logarithm , we can express \displaystyle \log_{2} and \displaystyle \log_{3} using ln,

\displaystyle \log_{2}x=\frac{\ln x}{\ln 2}\quad and \displaystyle \quad\log_{3}x = \frac{\ln x}{\ln 3}\,\textrm{.}

The two terms \displaystyle \log_3 118 and \displaystyle \log_3\log_2 3 can therefore be written as

\displaystyle \log_{3}118 = \frac{\ln 118}{\ln 3}\quad and \displaystyle \quad\log_{3}\log_{2}3 = \log_{3}\frac{\ln 3}{\ln 2}\,,

where we can simplify the last expression further with the logarithm law, log (a/b) = log a – log b, and then transform \displaystyle \log _{3} to ln,

\displaystyle \begin{align}

\log_{3}\frac{\ln 3}{\ln 2} &= \log_{3}\ln 3 - \log_{3}\ln 2\\[5pt] &= \frac{\ln\ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.} \end{align}

In all, we thus obtain

\displaystyle \log_{3}\log_{2}3^{118} = \frac{\ln 118}{\ln 3} + \frac{\ln \ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}

Input into the calculator gives

\displaystyle \log_{3}\log_{2}3^{118}\approx 4\textrm{.}762\,\textrm{.}


Note: The button sequence on the calculator will be:


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